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sukhopar [10]
3 years ago
10

Find the inequality represented by the graph.

Mathematics
1 answer:
anzhelika [568]3 years ago
7 0

Answer:

Step-by-step explanation:

y intercept is 3

y - 4 = 3 ( x - 4)

y - 4 = 3x - 12

y < 3x - 8

not equal to because the line is dashed.

not greater than because the shading is below the line

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|x - 2| + 10 = 12
|x - 2| = 12 - 10 = 2
x - 2 = 2 or x - 2 = -2
x = 2 + 2 or x = -2 + 2
x = 4 or x = 0
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Approximate the value of V110 to the nearest hundredth. HURRRYYYY PLZZZZZ NOWWWW​
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10.49

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What is the greatest common factor (GCF) of 81 and 18? A. 3 B. 81 C. 9 D. 162
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Answer:

C. 9

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Quick algebra 1 question for 10 points!
zaharov [31]

<u>Answer:</u>

○ x = -7

<u>Step-by-step explanation:</u>

Given the equation:

10(x + 10) - 4 = 11 - 5(2x + 11),

to solve for x, we have to rearrange the equation to make x its subject.

10(x + 10) - 4 = 11 - 5(2x + 11)

⇒ 10x + 100 - 4 = 11 - 10x -55      [expand brackets]

⇒ 10x + 96 = -44 - 10x

⇒ 10x + 10x + 96 = -44                   [add 10x to both sides]

⇒ 20x + 96 = -44

⇒ 20x = -44 - 96                             [subtract 96 from both sides]

⇒ 20x = -140

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In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A
Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let p_1 be the probability that a bearing meets the specification.

So, p_1=0.9

Sample size, n_1=500, is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: \mu_1=n_1p_1=500\times0.9=450

Variance: \sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45

\Rightarrow \sigma_1 =\sqrt{45}=6.71

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, X\geq 440.

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56.

So, the probability that a given shipment is acceptable is

P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062

Hence,  the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by p_2.

p_2=0.94

The total number of shipment, i.e sample size, n_2= 300

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, \mu_2, and variance, \sigma_2^2.

Mean: \mu_2=n_2p_2=300\times0.94=282

Variance: \sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92

\Rightarrow \sigma_2=\sqrt(16.92}=4.11.

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85.

So, the probability that a given shipment is acceptable is

P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977

Hence,  the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, \alpha be the required probability of acceptance of one shipment.

So,

-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}

On solving

\alpha= 0.977896

Again, the probability of acceptance of one shipment, \alpha, depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let p be the probability that one bearing meets the specification. So

-2.01=\frac{439.5-500  p}{\sqrt{500 p(1-p)}}

On solving

p=0.9053

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0
3 years ago
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